All categories

3 years ago

the highest possible score on an entrance exam is 800 points. what scores are between 80% to 90% of the highest possible score?

Mathematics

1 answer:

garik1379 [7]3 years ago

5 0

Answer: 640-720

Step-by-step explanation:

You might be interested in

Can someone please tell me the answer on this.

marishachu [46]

Answer: x = 31

Step-By-Step:

To Find x:

$\left(2x+20\right)+\left(x+10\right)+\left(2x-5\right)=180$ (Angle Sum Property Of Triangles)

(Remove the brackets)

$2x+20+x+10+2x-5=18$

(Group Accordingly)

$2x+x+2x+20+10-5=180$

= $5x+20+10-5=180$

= $5x\:+\:25$ $=180$

= $5x= 180 - 25$

= $5x = 155$

= $x=\frac{155}{5}$

= $31$

Therefore x = 31

So,

Angle A = x + 10 = 31 + 10 = 41

Angle B = 2*x + 20 = 2*31 + 20 = 82

Angle C = 2*x - 5 = 62 - 5 = 57

5 0

3 years ago

Who can solve this 2xy if x=5 and y=-3

Softa [21]

Answer:

-30

Step-by-step explanation:

2*5 is 10

10*-3 is -30

8 0

3 years ago

Read 2 more answers

At noon, the temperature was 5 degrees Celsius. By late afternoon, it has risen 6 degrees Celsius. What was the temperature late

KonstantinChe [14]

Answer:

11 degress celcius

Step-by-step explanation:

there was an original 5 but it increased by 6

5 + 6 = 11

11 degrees celcuis

7 0

3 years ago

Read 2 more answers

A metal ball of mass 100 g is heated to 90°C and then cooled to 25°C. The heat lost in the process is 2.5 kJ. Another metal ball

frosja888 [35]

specific heat ( $C_p 
$ ) - the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree)

the equation related to this theory is
 $Q=mC_p \Delta T$

Where Q is the energy gained or lost from the system
m is the mass of the object
 $C_p$ is the specific heat of the material
$\Delta T$ is the change in temperature

5 0

4 years ago

Read 2 more answers

What is the answer???

solmaris [256]

$Method\ 2:\\\\a_n=a_{n-1}+\dfrac{1}{2}\\\\\text{It's an arithmetic sequencion where}\ a_1=-\dfrac{3}{2}\ \text{and common difference}\ d=\dfrac{1}{2}.\\\\\text{The formula of n-th term is}\ a_n=a_1+(n-1)d.\ \text{Substitute}\\\\a_n=-\dfrac{3}{2}+(n-1)\left(\dfrac{1}{2}\right)=-\dfrac{3}{2}+\dfrac{1}{2}n-\dfrac{1}{2}=-\dfrac{4}{2}+\dfrac{1}{2}n=\dfrac{1}{2}n-2\\\\a_9=\dfrac{1}{2}(9)-2=4.5-2=2.5=\dfrac{5}{2}$ $a_1=-\dfrac{3}{2}\\\\a_n=a_{n-1}+\dfrac{1}{2}\\\\Method\ 1:\\\\a_2=a_1+\dfrac{1}{2}\to a_2=-\dfrac{3}{2}+\dfrac{1}{2}=-\dfrac{2}{2}\\\\a_3=a_2+\dfraC{1}{2}\to a_3=-\dfrac{2}{2}+\dfrac{1}{2}=-\dfrac{1}{2}\\\\a_4=a_3+\dfrac{1}{2}\to a_4=-\dfrac{1}{2}+\dfrac{1}{2}=0\\\\a_5=a_4+\dfrac{1}{2}\to a_5=0+\dfrac{1}{2}=\dfrac{1}{2}\\\\a_6=a_5+\dfrac{1}{2}\to a_6=\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{2}{2}\\\\a_7=a_6+\dfrac{1}{2}\to a_7=\dfrac{2}{2}+\dfrac{1}{2}=\dfrac{3}{2}$

$a_8=a_7+\dfrac{1}{2}\to a_8=\dfrac{3}{2}+\dfrac{1}{2}=\dfrac{4}{2}\\\\a_9=a_8+\dfraC{1}{2}\to a_9=\dfrac{4}{2}+\dfrac{1}{2}=\dfrac{5}{2}$

5 0

4 years ago